Linear and continuous applications (normed space)

Question

Hello can you help me with this exercise i don't ununderstand how I must do

We provide the vectorial space $\mathbb{R}^2$ with the norm $$ \|(x,y)\|= \sqrt{\frac{x^2}{16}+\frac{y^2}{9}} $$ prove that $u(x,y)=\sqrt{2}x+y$ is linear and continuous on $\mathbb{R}^2$

the linearity is clear but how to do for the continuity?

I must prove that $$ \|u\|\leq c $$ where $$\|u\|= \sup_{\|(x,y)\|\leq 1} \|u(x,y)\|$$ how to do? Thank you for your help

Answer 1

Hint:

$\|u(x,y)\|=\|\sqrt{2}x+y\|=|\sqrt{2}x+y|$ (the standard norm on $\mathbb{R}$).

So try to maximize the function $f(x,y)=|\sqrt{2}x+y|$ over the elliptical region $\frac{x^2}{16}+\frac{y^2}{4} \leq 1$.

Answer 2

For the continuity let $(x_n,y_n)$ be a sequence in $\mathbf R^2$ such that $(x_n,y_n) \to (x,y)$. All you have to do is to show that $$|u(x_n,y_n) - u(x,y)| = | \sqrt{2} (x_n-x) +y_n - y|\to 0, \qquad \text{for} \quad n \to +\infty.$$ Hint: What can you say about the convergence of the component sequences $(x_n)$ and $(y_n)$ in $\mathbf R$ if you know that $(x_n,y_n)$ converges in $\mathbf R^2$? Use the triangle inequality.